# Tag Info

13

Another term is thermal resistance, This is incorrect. Thermal resistance is something that prevents heat flow. It is an entirely separate concept from electrical resistance. How is contact resistance explained? To obtain very low resistance in a material like most metals, the electrons must be delocalized from the individual atoms, and free flow ...

5

This is a very interesting question, especially considering the very recent history of scholarship on electrical contact resistance (a term first coined in 1964 by William Shockley, one of the inventors of the transistor), as well as thermal contact resistance. For the following explanation, I will use this research paper on electrical contact resistance ...

3

Assume you have four waterfalls, and they connect one river to another. The waterfalls drop by five feet. If you put the waterfalls end to end, then the second river must be 20 feet below the first river. If, however, the waterfalls get put side by side, and still connect the two rivers, the second river is just five feet below the first river. Now, this ...

3

As you slide the switch open, it doesn't instantly transition from zero resistance to infinite resistance. As the contact area decreases the resistance rises, which acts upon the current to produce a voltage operating against the current. Even when the contact resistance becomes zero, there is capacitance across the gap, which produces a rapidly rising ...

2

Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current. If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the ...

2

This is more of a math question, but it's just a trigonometric identity. $$\cos\Big(\frac{\pi}{2}-x\Big)=\sin(x)\\$$ Or $$\cos\Big(x+\frac{\pi}{2}\Big)=-\sin(x)$$

2

The temperature of the circuit can probably be crudely modeled by assuming that you have two channels: one for heat absorption, and one for heat loss. Heat loss Suppose you have something like a light bulb. The hot filament loses heat via radiation and conduction. Let's focus on radiation. The rate of heat radiated by an object of temperature $T$ depends ...

2

When one says that bulb is 100-W, does that mean it is 100-W at 120V, which would tell me the resistance of the bulb? Somehow I have to find the resistance of the bulb which is not given. Recall that, for a resistor, the AC power dissipated is $$P_R = \frac{v_{rms}^2}{R}$$ Assuming the voltage across the bulb is not significantly reduced by the ...

2

Electrons do not "decide" which path to take in any meaningful precise sense (they don't take any particular path at all unless an interaction fixing their position takes place every step along the way), hence there is no time span in which that decision is made.

2

I guess the answer you are looking for is that the electric field propagates at the speed of light. Suddenly add a voltage source to a complete circuit and the electric field will spread at the speed of light $c$. Depending on how far away a specific electron is in the circuit, this electron will soon feel this electric field and then immediately react to ...

1

The current will be the same in the wire and the resistor if they are in series. This is ensured by conservation of charge - you cannot have more electrons leaving a component than entering it. (Kirchoff's current law)

1

For a truly ideal conductor, the voltage is identical at all points on the conductor in a steady state. So if you attached this to a real battery, the battery would shove charge along "trying" to maintain the voltage between the terminals. It wouldn't be able to do this. A very powerful battery might heat up trying to do this so much it damages itself or ...

1

With a perfect conductor, there is no resistance in the wire. However, that does not mean there is no resistance in the circuit, as the battery itself has some internal resistance. The current through the wire will then be limited by that resistance, i.e. $$I=E/R_i$$ The internal resistance is quite low. It has to be as otherwise its voltage would drop too ...

1

In simple circuit is current in wire different from current in resistor as wire also opposes the flow of electrons so wire should also be added in series connection of resistor IF you follow this answer step by step the situation should be MUCH clearer to you. IF you just skim this answer it will completely confuse you. Short summary: Read my ...

1

A magnetic field is determined by the current and a changing electric field. And it has energy just for existing. It takes energy to make the magnetic field, for instance to increase the current, and you get energy back when magnetic fields decrease in strength. For a common inductor the magnetic field and associated stored energy are due solely to the ...

1

The unit Wh, shown on the battery, is Watt hours, which is the total energy (Joules), not the power.

1

You should look at an electric circuit a like a river or a water slide attraction in an amusement park (see my little artist impression below). The resistors are the steep parts: that's where the potential energy is lost. The wires are the horizontal parts, so there no potential energy is lost. But as the water is already moving, it doesn't stop moving in ...

1

You're exactly in trouble! Wh is a symbol of ENERGY, not the POWER!

1

The main requirements are that Current flow into or out of "static sensitive" components be limited to below a level where damage is caused by I^2 x R heating. Devices which are damaged by voltage without current flow (such as Field Effect Transistors gate to drain voltages) do not have excessive differential voltage applied. If all "bodies" concerned ...

1

First, do not put them "in series"! Each is powered from the mains ie they are in parallel. Second, earthing is typically applied to any external metalwork that anyone might come into contact where there is any danger of the live coming loose and touching that metal. Think worst case fault conditions. Finally, the fact you have to ask these question means ...

1

I can verify using Kirchoff's laws that this is solvable. Ultimately, $$R_{equiv} = \frac{I_aa+I_bb}{I_a+I_e}$$ where the subscripts denote which resistor the current flows through. Setting up $I_e = I_d+I_c$, $I_b = I_a+I_c$, $I_ee + I_cc - aI_a = 0$, $I_dd - bI_b = cI_c$, which is 5 unknowns, but only 4 equations. This isn't a problem, as you can soon ...

1

I don't think there is a "preferred" type of LC circuit. How to wire your circuit depends on how you want your circuit to work. Notably the impedance is different in the two cases, while the resonant frequency is the same according to Wikipedia. If you have a look at the equations they present you can quickly see that in terms of what the two kinds circuits ...

1

As stated in the comments, any loss in the system will contribute to the Johnson noise, so you are right about the skin effect and the Eddy current. I want to add that, interestingly enough, this apply not only to electric circuits, but to other linear dissipative systems. A very interesting paper from 1951, Irreversibility and Generalized Noise, proves it ...

1

When a motor moves it also acts as a generator and the current trough the windings is given by the difference of the external voltage and the induced voltage. When the motor stands still, though, the generated voltage is zero and the windings will draw the max. current they can based on their DC resistance. In other words, the faster the motor runs, the ...

1

The standard trick is to split off the circuit after the first link, and treat the 'tail' as another copy of the circuit itself. This means that the impedance $Z$ of the whole circuit must satisfy $$Z=2Z_1+\frac{1}{\frac{1}{Z_2}+\frac{1}{Z}}.$$ This gives a quadratic equation in $Z$ which is easy enough to solve.

1

Always remember that you are free to deform and stretch a circuit without changing its topological properties, i.e. no deleting or adding nodes, or popping them over circuit elements. The easiest way to see that R1, R2, and R3 are all in parallel is to pull R1 to the left along its wire until it is vertical, and similarly pull R2 to the right along its wire ...

1

OK. Let us start with the inital equation $$Z=Z_1\left[1\pm\sqrt{1+2Z_2/Z_1}\right] ;\ (1)$$ and consider lossless circuit. It is a good idea to determine the value of $Z_1$ as $Z_1=ix_1$ and similarly, $Z_2=ix_2$ where $x$ denotes the reactance. For the the capacitive reactance $x<0$ and for the inductive reactance $x>0$. We have: ...

1

I suppose that in this type of circuit the current and the voltage are in phase, This question is asking for a transient solution, not an ac steady-state solution. So we can't really talk about the phase of the voltage or the current. meaning that immediately after the switch is shut the current should be I=0? This conclusion is incorrect. ...

1

Why does it maintain the status quo? There is energy stored in the magnetic field, and the magnetic field is proportional to the current. In order for the current to change, the magnetic field must change. By conservation of energy, that means the magnetic field energy must be transformed to some other kind of energy. Until you provide a mechanism ...

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